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Article

On the Domain of the Fibonacci Difference Matrix

1
Kilis Yatırım Destek Ofisi, Şehitler Mah. Cambazlar Sok. No:9, Kilis 79000, Turkey
2
Faculty of Arts and Sciences, Department of Mathematics, Gaziantep University, Gaziantep 27310, Turkey
*
Author to whom correspondence should be addressed.
Mathematics 2019, 7(2), 204; https://doi.org/10.3390/math7020204
Submission received: 31 December 2018 / Revised: 17 February 2019 / Accepted: 19 February 2019 / Published: 21 February 2019

Abstract

:
Matrix F^ derived from the Fibonacci sequence was first introduced by Kara (2013) and the spaces lp(F) and l(F); (1 ≤ p < ∞) were examined. Then, Başarır et al. (2015) defined the spaces c0(F) and c(F) and Candan (2015) examined the spaces c(F(r,s)) and c0(F(r,s)). Later, Yaşar and Kayaduman (2018) defined and studied the spaces cs(F(s,r)) and bs(F(s,r)). In this study, we built the spaces cs(F) and bs(F). They are the domain of the matrix F on cs and bs, where F is a triangular matrix defined by Fibonacci Numbers. Some topological and algebraic properties, isomorphism, inclusion relations and norms, which are defined over them are examined. It is proven that cs(F) and bs(F) are Banach spaces. It is determined that they have the γ, β, α -duals. In addition, the Schauder base of the space cs(F) are calculated. Finally, a number of matrix transformations of these spaces are found.

1. Introduction

Cooke [1] formulated the theory of infinite matrices in the book “Infinite Matrices and Sequence Spaces”. Many researchers have investigated infinite matrices after the publication of this book in 1950. In most of these studies, the domain of infinite matrices on a sequence space was studied. In this study, we address the question: What are the properties of the domain of the Fibonacci band matrix on sequence spaces bs and cs? The domain of the Fibonacci band matrix creates a new sequence space. We handle algebraic properties of this new space in order to determine its duals and its place among other known spaces, and to characterize the matrix transformations of this space.
One difficulty of this study is to determine whether the new space is the contraction or the expansion, or the overlap of the original space. Another difficulty is to determine the matrix transformations on this space and into this space. For the first problem, we give a few inclusion theorems. For the second problem, we use the matrix transformation between the standard sequence spaces and two theorems.
Generating a new sequence space and researching on its properties have been important in the studies on the sequence space. Some researchers examined the algebraic properties of the sequence space while others investigated its place among other known spaces and its duals, and characterized the matrix transformations on this space.
We can create a new sequence space by using the domain of infinite matrices. Ng-Lee [2] first investigated the domain of an infinite matrix in 1978. In the same period, Wang [3] created a new sequence space by using another infinite matrix. Many researchers such as Malkovsky [4], Altay, and Başar [5] followed these studies. This topic was studied intensively after 2000.
Leonardo Fibonacci invented Fibonacci numbers. He introduced Fibonacci numbers originated from a rabbit problem. These numbers create a number sequence:
1,1,2,3,5,8,13,21,34,55,89,....
This sequence has important properties and applications in various fields.
Let us indicate the Fibonacci sequence by (fn). fn is defined as
fn = fn−1 + fn−2 (n ≥ 2)
with f0 = f1 = 1. The golden ratio is
lim n f n + 1 f n = 1 + 5 2 = ϕ .
Let us indicate the set of all real-valued sequences with w and list some subspaces of w called standard sequence spaces.
c = { r = ( r k ) w : lim k | r k p | = 0     for   some   p } ,
c 0 = { r = ( r k ) w : lim k | r k | = 0 } ,
b s = { r = ( r k ) w : sup n | k = 0 n r k | < } , c s 0 = { r = ( r k ) w : lim n | k = 0 n r k | = 0 } , l = { r = ( r k ) w : sup k | r k | < } ,
b v = { r = ( r k ) w : k = 0 | r k r k 1 | < } ,
b v 0 = b v c 0 l p = { r = ( r k ) w : k = 0 | r k | p < ,    0 < p < } , l 1 = { r = ( r k ) w : k = 0 | r k | < } ,
Now let us take real valued infinite matrix T = (tnk), where tnk is a real number for every n,k∈ℕ Let A and B be sequence spaces. Sequence Tx = {Tn(x)} is T-transform of a for every a = (ak)∈A. Here, TaB and
T n ( a ) = k t n k a k
and Tn(a)t (t exists for every n∈ℕ). Then, T is called a matrix transformation from A to B.
Now let us take infinite matrix T and sequence space δ to define domain of infinite matrix T. The domain of the matrix T on δ is characterized by
δT = {x = (xk)∈w: Txδ}.
Many reserachers have studied the domain of a matrix on a sequence space. For more detailed information on these new sequence spaces, see references [6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26].
The Fibonacci difference matrix F was first introduced by Kara [27] in 2013. F is derived from (fn). In this study, Kara [27] defined the spaces lp(F) and l(F); (1 ≤ p < ∞). After this study the c0(F(r,s)) and c(F(r,s)) was introduced by Candan [28], in 2015, where F(r,s) is a generalized Fibonacci matrix. Candan and Kara [19] introduced and examined lp(F(r,s)); (1 ≤ p < ∞). In 2018, Yaşar and Kayaduman [29] defined and examined cs(F(s,r)) and bs(F(s,r)) and Kayaduman and Yaşar [30] studied spaces bs(Nt) and cs(Nt), where Nt is a Nörlund matrix.
Let δ be a sequence space. The γ, β, α -duals of δ are defined, respectively, as follows
δ γ = { x = ( x k ) w : x s = ( x k s k ) b s   for   all   s δ } ,
δ β = { x = ( x k ) w : x s = ( x k s k ) c s   for   all   s δ } ,
δ α = { x = ( x k ) w : x s = ( x k s k ) l 1   for   all   s δ } .
In this study, spaces cs(F) and bs(F) are introduced and the related notations are given in Section 2. In addition, some topological and algebric properties, isomorphism, inclusion relations and norms which are defined over them are examined. The γ, β, α -duals of these spaces are determined in Section 3. The Schauder base of space cs(F) are calculated. Finally, many matrix transformations of these spaces are found. In the last section, the results and previous studies and the working hypotheses are discussed.
A detailed literature review was performed before this study was started. Scans were made on related articles, magazines, and books. As a result of these scans, the part related to our subject was synthesized and the results were noted. These results were then applied to our problem area. Finally, the results of this study were obtained.

2. Results

2.1. The Domain of Fibonacci Difference Matrix F on Bounded and Convergent Series

In this section, cs(F) and bs(F) are introduced. Related notations are given. In addition, some topological and algebric properties, isomorphism, inclusion relations, and norms defined over them are examined.
For similar studies, see refeerences [19] and [27,28,29,30,31,32,33,34].
Let spaces cs(F) and bs(F) be the domain of the matrix F on cs and bs, where F = {fnk} infinite matrix is defined by (fn)
f n k = { f n + 1 f n , k = n 1 f n f n + 1 , k = n 0 , 0 k < n 1 or   n < k
for all k,n∈ℕ. Then we inroduce cs(F) and bs(F) as
c s ( F ) = { x = ( x k ) w : ( k = 0 n ( f k f k + 1 x k f k + 1 f k x k 1 ) ) n c }
b s ( F ) = { x = ( x k ) w : sup n | k = 0 n ( f k f k + 1 x k f k + 1 f k x k 1 ) | < } .
We can see cs(F) = (cs)F and bs(F) = (bs)F by using Equation (2).
Let the inverse matrix of F be F−1. For all k,n∈ℕ, F−1 = {F−1nk} is found as
f n k 1 = { f n + 1 2 f k f k + 1 , 0 k n 0 , k > n .
Let us take sequence x = (xn). If y = Fx, then we calculate as
y n = ( F x ) n = { x 0 ,        n = 0 f n f n + 1 x n f n + 1 f n x n 1 ,       n 1 .
Herefrom, if we calculate inverse of F, then we find that x = F−1y and
x n = k = 0 n f n + 1 2 f k f k + 1 y k .
Now, let us give some theorems related to our study.
Theorem 1.
bs(F) is a linear space.
Proof. 
The proof is left to the reader since it is easy to show. □
Theorem 2.
cs(F) is a linear space.
Proof. 
The proof is left to the reader since it is easy to show. □
Theorem 3.
bs(F) is a normed space with:
x = sup n | j = 0 n ( f k f k + 1 x k f k + 1 f k x k 1 ) |
Proof. 
The proof is left to the reader since it is easy to show. □
Theorem 4.
cs(F) is a normed space with Equation (6).
Proof. 
The proof is left to the reader since it is easy to show. □
Theorem 5.
bs(F) is isomorphic to bs.
Proof. 
Let us take T: bs(F) →bs mentioned Equation (4) by xy = Tx = Fx. It is easy to see that T is linear and injective.
We must find T is surjective. Let y = (yn)∈bs. By using Equation (5) and Equation (6), we see
x = sup n | k = 0 n ( f k f k + 1 x k f k + 1 f k x k 1 ) | = sup n | k = 0 n ( f k f k + 1 ( j = 0 k f k + 1 2 f j f j + 1 y j ) f k + 1 f k ( k = 0 k 1 f k + 1 2 f j f j + 1 y j ) ) | = sup n | k = 0 n y k | = y b s < .
We see that xbs(F). Hence, T is surjective. In addition, bs(F) and bs izometric because x b s ( F ) = y b s .
Theorem 6.
cs(F) is isomorphic to cs.
Proof. 
The proof can be made similar to Theorem 5, so it is left to the reader. □
Theorem 7.
bs(F) is a Banach space with Equtaion (6).
Proof. 
It is easy to see the norm conditions are ensured. Let a Cauchy sequence xi = (xki) in bs(F) for each i∈ℕ. For all k, we have
y k i = f k f k + 1 x k i f k + 1 f k x k 1 i
from Equation (4). For all ε > 0 there is n0 = n0(ε) such that
x i x m b s ( F ) = sup n | k = 0 n ( f k f k + 1 ( x k i x k m ) f k + 1 f k ( x k 1 i x k 1 m ) ) | = sup n | k = 0 n ( y k i y k m ) | = y i y m b s < ε
for all i,m ≥ n0. yiy (i→∞) such that y b s exists, since b s is complete. Since bs and bs(F) are isomorphic, bs(F) is complete. It hereby is a Banach space. □
Theorem 8.
cs(F) is a Banach space with Equation (6).
Proof. 
It is easy to see the norm conditions are ensured. Let a Cauchy sequence xi = (xki) in cs(F) for each i∈ℕ. For all k, we have
y k i = f k f k + 1 x k i f k + 1 f k x k 1 i
from Equation (4). For all ε > 0, there is n0 = n0(ε) such that
x i x m c s ( F ) = sup n | k = 0 n ( f k f k + 1 ( x k i x k m ) f k + 1 f k ( x k 1 i x k 1 m ) ) | = sup n | k = 0 n ( y k i y k m ) | = y i y m c s < ε
for all i,m ≥ n0. yiy (i→∞) such that y c s exists, since c s is complete. Since cs and cs(F) are isomorphic, cs(F) is complete. It hereby is a Banach space. □
Now, let R = (rnk) infinite matrix. Let us list the following:
sup n k | r n k | < ,
lim k r n k = 0   for   each   n ,
sup m k | n = 0 m ( r n k r n , k + 1 ) | < ,
lim n k r n k = p   for   each   k N ,   p ,
sup n k | r n k r n , k + 1 | < ,
lim n r n k = a k   for   each   k ,   a k ,
sup N , K | n k ( r n k r n , k + 1 ) | < ,
sup N , K | n k ( r n k r n , k 1 ) | < ,
lim n ( r n k r n , k + 1 ) = a   for   each   k ,   a ,
lim n k | r n k r n , k + 1 | = k | lim n ( r n k r n , k + 1 ) | ,
sup n | lim k r n k | < ,
lim n k | r n k r n , k + 1 | = 0   uniformly   in   n ,
lim m k | n = 0 m ( r n k r n , k + 1 ) | = 0 ,
lim m k | n = 0 m ( r n k r n , k + 1 ) | = k | n ( r n k r n , k + 1 ) | ,
sup N , K | n k [ ( r n k r n , k + 1 ) ( r n 1 , k r n 1 , k + 1 ) ] | < ,
sup m | lim k n = 0 m r n k | < ,
a k n r n k = a k   for   each   k ,
sup N , K | n k [ ( r n k r n 1 , k ) ( r n , k 1 r n 1 , k 1 ) ] | < .
The collection of all finite subsets of ℕ denoted by .
Lemma 9.
Let us suppose infinite matrix R = (rnk). Then,
(1) R = ( r n k ) ( b s , l ) iff Equations (11) and (8) hold [35].
(2) R = ( r n k ) ( c s , c ) iff Equations (11) and (12) hold [36].
(3) R = ( r n k ) ( b s , l 1 ) iff Equations (13) and (8) hold [37].
(4) R = ( r n k ) ( c s , l 1 ) iff Equation (14) holds [35].
(5) R = ( r n k ) ( b s , c ) iff Equations (8), (15) and (16) hold [37].
(6) R = ( r n k ) ( c s , l ) iff Equations (17) and (11) hold [35].
(7) R = ( r n k ) ( b s , c 0 ) iff Equations (18) and (8) hold [35].
(8) R = ( r n k ) ( b s , c s 0 ) iff Equations (19) and (8) hold [38].
(9) R = ( r n k ) ( b s , c s ) iff Equations (20) and (8) hold [38].
(10) R = ( r n k ) ( b s , b v ) iff Equations (21) and (8) hold [38].
(11) R = ( r n k ) ( b s , b s ) iff Equations (8) and (9) hold [38].
(12) R = ( r n k ) ( c s , c s ) iff Equations (9) and (10) hold [39].
(13) R = ( r n k ) ( b s , b v 0 ) iff Equations (21), (18) and (21) hold [35].
(14) R = ( r n k ) ( c s , c 0 ) iff Equations (11) and (12) hold with ak = 0 for all k ∈ ℕ [40].
(15) R = ( r n k ) ( c s , b s ) iff Equations (9) and (22) hold [38].
(16) R = ( r n k ) ( c s , c s 0 ) iff Equations (9) and (23) hold with ak = 0 for all k ∈ ℕ [38].
(17) R = ( r n k ) ( c s , b v ) iff Equation (24) holds [38].
(18) R = ( r n k ) ( c s , b v 0 ) iff Equations (24) and (12) hold with ak = 0 for all k ∈ ℕ [35].
Theorem 10.
bs(F) bs is valid.
Proof. 
Suppose xbs. If we show that F is an element of (bs,bs) then x is element of bs(F). For this, F must provide Equations (8) and (9). Since lim k f n k = 0   for   each   n , Equation (8) is provided.
If we examine Equation (9), we find
sup m k | n = 0 m ( f n k f n , k + 1 ) | = 0 .
Theorem 11.
bs(F) is not valid.
Proof. 
Suppose x = (xk) = (f2k+1). Then y = Fx = (1,0,0....)∈bs. On the other hand, f2k+1 as k→. It is clear xbs(F), but x. This result completes the proof. □
Theorem 12.
cs(F) cs is valid.
Proof. 
Suppose xcs. If we show that F is element of (cs,cs) then x is element of cs(F). For this, F must provide Equations (10) and (9). Equation (9) has been provided from the Theorem 10. If we look at the Equation (10) then, for each k,
lim n k f n k = lim n ( f n f n + 1 f n + 1 f n ) = 1 ϕ ϕ = l
such that l exists. □
Theorem 13.
cs(F) c is not valid.
Proof. 
Let x = (xk) = (f2k+1). Then y = Fx = (1,0,0....)∈cs. On the other hand, f2k+1 as k→. It is clear xcs(F), but xc. This result completes the proof. □
Theorem 14.
cs(F) ⊂ bs(F) is valid.
Proof. 
If xcs(F), y = Fxcs. Hence, k F x c . Since c, k F x l . Hence, Fxbs. That is, xbs(F). This result completes the proof. □
Let us take normed space A and let (ak) A. If there is only one scalar sequence (vk) such that y = k = 0 ν k a k and lim n y k = 0 n ν k a k = 0 then (ak) is called a Schauder base for A.
Now, let us give corallary releated to Schauder basis.
Corallary 15.
Let a sequence u ( k ) = { u n ( k ) } n in cs(F) be for each k ∈ ℕ and
u n ( k ) = { f n + 1 2 f k f k + 1 , 0 k n 0 , n < k .
Then { u n ( k ) } n is a base for cs(F). Every xcs(F) can write as a single x = k = 0 y k u k such that y k = ( F ^ x ) k .

2.2. The Duals of cs(F) and bs(F) and Matrix Transformations

Let us give the two lemmas to use in the next stage.
Lemma 16.
Let infinite matrix C = ( c n k ) 1 2 and a = ( a n ) w . Let us take C = a F 1 , that is,
c n k = { a n f n k 1 , 0 k n 0 , n < k
for all k , n , δ { c s , b s } . Then, a { δ ( F 1 ) } α iff C { δ , l 1 } .
Proof. 
Let x = (xn) and a = (an) elements of w. y = (yn) such that y = Fx which is defined in Equation (4). If we use to Equation (4), then
a n x n = a n ( F 1 y ) n = ( C y ) n .
ax = (an xn)∈₁ with x = (xn)∈µ(F) iff Cy₁ with yλ. Consequently, C(µ,ℓ). □
Lemma 17.
[41]Let us take a = (ak)∈w and infinite matrix C = (cnk). Let the inverse matrix H = (hnk) of the triangular matrix G = (gnk) is given by
c n k = { j = k n a j h j k , 0 k n 0 , n < k .
Then, for any sequence space δ,
δ G γ = { a = ( a k ) w : C ( δ , l )   } ,
δ G β = { a = ( a k ) w : C ( δ , c )   } .
If we consider Lemma 9, Lemma 16 and Lemma 17 together, the following is obtained;
Corallary 18.
Let us take r = (rk)∈ w and infinite matrix A = (ank) and B = (bnk) such that
a n k = { r n f n k 1 , 0 k n 0 , n < k and   b n k = j = k n f j + 1 2 f k f k + 1 r j .
If we take d1, d2, d3, d4, d5, d6, d7 and d8 as follows:
  • d 1 = { r = ( r k ) w : sup N , K | n k ( a n k a n , k + 1 ) | < } , d 2 = { r = ( r k ) w : sup N , K | n k ( a n k a n , k 1 ) | < } , d 3 = { r = ( r k ) w : lim k c n k = 0 } , d 4 = { r = ( r k ) w : α lim n ( b n k b n , k + 1 ) = α   f o r a l l k } ,
  • d 5 = { r = ( r k ) w : lim n k | b n k b n , k + 1 | = k | lim n ( b n k b n , k + 1 ) |   } , d 6 = { r = ( r k ) w : α lim n b n k = α    f o r a l l k   } , d 7 = { r = ( r k ) w : sup n k | b n k b n , k + 1 | < } , d 8 = { r = ( r k ) w : sup n | lim k b n k | < } .
then,
(1) {bs(F)}α = d1
(2) {cs(F)}α = d2
(3) {bs(F)}β = d3∩d4∩d5
(4) {cs(F)}β = d6∩d7
(5) {bs(F)}γ = d3∩d7
(6) {cs(F)}γ = d7∩d8.
Theorem 19.
Let μ { c s , b s } and λ ⊂ w. Then, U = ( u n k ) ( μ ( F ) , λ ) iff
V m = ( v n k ( m ) ) ( μ , c )   for   all   n
V = ( v n k ) ( μ , λ ) ,  
where
v n k ( m ) = { j = k m f j + 1 2 f k f k + 1 u n j , 0 k m 0 , m < k
and
v n k = j = k f j + 1 2 f k f k + 1 u n j
for all k , m , n .
Proof. 
Necessity part: Let us take that A = ( a n k ) ( μ ( F ) , λ ) and x = ( x k ) μ ( F ) . If we use Equation (5), then we find
k = 0 m a n k x k = k = 0 m a n k k = 0 m f j + 1 2 f k f k + 1 y j = k = 0 m j = k m f j + 1 2 f k f k + 1 a n k y k = k = 0 m d n k ( m ) y k = D n ( m ) ( y ) .
According to the hypothesis, for each m∈ℕ, Am(x) ∈ c. Then, V ( m ) c for each m∈ℕ and V ( m ) ( μ , c ) . Ax = Vy if we consider for m from Equation (30). As a result, we find V = ( v n k ) ( μ , λ ) .
Sufficient part: Suppose that Equations (26) and (27) are satisfied and x = ( x k ) μ ( F ) be. By using Corollary 18 and Equations (26) and (30), we have that y = F x μ and
V n ( m ) ( y ) = k = 0 m v n k ( m ) y k = k = 0 m a n k x k = A n ( m ) ( x ) c .
Hence, A = ( a n k ) k exists. Ax = Vy if we consider for m→ from Equation (30). Consequently, we find A = ( a n k ) ( μ ( F ) , λ ) .
Theorem 20.
Let μ ∈ {bs,cs} and λ ⊂ w and U = (unk) and B = (bnk) be infinite matrices. Let
b n k : = f n f n + 1 u n k f n + 1 f n u n 1 , k .
Then, U ( λ , μ ( F ^ ) ) iff B ( λ , μ ) .
Proof. 
Let z = (zk)∈ l and Equation (31) exist. Then, we have
k = 0 m b n k z k = k = 0 m ( f n f n + 1 a n k f n + 1 f n a n 1 , k ) z k .
If we take m→ to Equation (32), we have that (Bz)n = (F(Az))n. Consequently, Azμ(F) iff Bzμ. That is, B ( λ , μ ) . □
Let us give almost convergent sequences space, which was first defined by Lorentz [42]. Let t = (tk). t is almost convergent to limit iff lim m k = 0 m t n + k m + 1 = α uniformly in n. It is denoted by c ^ lim t = α . In addition, c ^ s and c ^ 0 mean the spaces of almost convergent series and almost null sequences, respectively. c ^ 0 and c ^ are
c ^ 0 = { x = ( x k ) l : lim m k = 0 m x n + k m + 1 = 0   uniformly   in   n } , c ^ = { x = ( x k ) l : l lim m k = 0 m x n + k m + 1 = α   uniformly   in   n } .
Now, let us take infinite matrix R = (rnk) and list the following:
a k f lim r n k = a k   for   each   k ,
lim q k 1 q + 1 | i = 0 q Δ [ j = 0 n + i ( r j k a k ) ] | = 0   uniformly   in   n ,
sup n k | Δ [ j = 0 n r j k ] | < ,
a k f lim j = 0 n r j k = a k   for   each   k ,
sup n k | j = 0 n r j k | < ,
a k n k r n k = a k   for   each   k ,
lim n k | Δ [ j = 0 n ( r j k a k ] | = 0 ,
sup n k | j = 0 n r j k | p < ,   q = p p 1 ,
sup m , n | n = 0 m r n k | < ,
sup m , l | n = 0 m k = l r n k | < ,
sup m , l | n = 0 m k = 0 l r n k | < ,
lim m k | n = m r n k | = 0 ,
n k r n k   convergent
lim m n = 0 m ( r n k r n , k + 1 ) = a   for   each   k ,   a .
Lemma 21.
Let infinite matrix R = (rnk) for all k,n∈ℕ. Then,
(1) R = ( r n k ) ( c ^ , c s ) iff Equation (23) and Equations (37)–(39) hold [43].
(2) R = ( r n k ) ( c s , c ^ ) iff Equation (11) and Equation (33) hold [44].
(3) R = ( r n k ) ( b s , c ^ ) iff Equations (8), (11), (33) and (34) hold[45].
(4) R = ( r n k ) ( b s , c ^ s ) iff Equations (8) and (34)–(36) hold[45].
(5) R = ( r n k ) ( c s , c ^ s ) iff Equation (35) and Equation (36) hold[44].
(6) R = ( r n k ) ( l , b s ) = ( c , b s ) = ( c 0 , b s ) iff Equation (37) holds[38].
(7) R = ( r n k ) ( l p , b s ) iff Equation (40) holds [46].
(8) R = ( r n k ) ( l , b s ) iff Equation (41) holds [38].
(9) R = ( r n k ) ( b v , b s ) iff Equation (42) holds [38].
(10) R = ( r n k ) ( b v 0 , b s ) iff Equation (43) holds [46].
(11) R = ( r n k ) ( l , c s ) iff Equation (44) holds [38].
(12) R = ( r n k ) ( c , c s ) if and only Equations (10), (37) and (45) hold [38].
(13) R = ( r n k ) ( c s 0 , c s ) iff Equations (9) and (46) hold [38].
(14) R = ( r n k ) ( l p , c s ) iff Equations (10) and (40) hold [46].
(15) R = ( r n k ) ( l , c s ) iff Equations (10) and (41) hold [46].
(16) R = ( r n k ) ( b v , c s ) if and only Equations (10), (41) and (43) hold [38].
(17) R = ( r n k ) ( b v 0 , c s ) iffEquations (10) and (43) hold [46].
Now, suppose v n k and v n k ( m ) which mentioned Equations (28) and (29) and give the following equations
lim k v n k ( m ) = 0   for   each   n ,
lim n k | v n k ( m ) v n , k + 1 ( m ) | = 0   uniformly   in   n ,
v k lim n   ( v n k ( m ) v n , k + 1 ( m ) ) = v k   for   each   k ,
lim k v n k = 0   for   each   n ,
sup n k | v n k v n , k + 1 | < ,
v k lim n   ( v n k v n , k + 1 ) = d k   for   each   k ,
l lim n k | v n k v n , k + 1 | = l   uniformly   in   n ,
sup m k | n = 0 m ( v n k v n , k + 1 ) | < ,
lim m k | n = 0 m ( v n k v n , k + 1 ) | = 0 ,
lim m k | n = 0 m ( v n k v n , k + 1 ) | = k | n ( v n k v n , k + 1 ) | = 0 ,
sup N , K | n k [ ( v n k v n , k + 1 ) ( v n 1 , k v n 1 , k + 1 ) ] | < ,
sup N , K | n k ( v n k v n 1 , k ) | < .
v k lim n   v n k ( m ) = v k   for   each   k ,
sup n k | v n k ( m ) v n , k + 1 ( m ) | < ,
sup n | lim k v n k | < ,
v k lim n   v n k = v k   for   each   k ,
sup m | lim k n = 0 m v n k | < ,
sup m k | n = 0 m ( v n k v n , k 1 ) | < ,
v k n v n k = v k   for   each   k ,
sup N , K n | k ( v n k v n , k 1 ) | < ,
v k f lim v n k = v k   for   each   k ,
sup N , K | n k [ ( v n k v n 1 , k ) ( v n , k 1 v n 1 , k 1 ) ] | < ,
lim q k 1 q + 1 | i = 0 q Δ [ j = 0 n + i ( v j k l k ) ] | = 0   uniformly   in   n ,
sup n k | j = 0 n v j k | < ,
l n k v n k = l  
lim n k | Δ [ j = 0 n ( v j k l k ] | = 0 ,
sup n k | Δ [ j = 0 n v j k ] | < ,
v k f lim j = 0 n v j k = v k   for   each   k .
If we consider Theorems 19 and 20, Lemmas 9 and 21, then we give the below conclusions.
Corollary 22.
Let us take U = ( u n k ) mentioned in Theorem 19. Then,
(1) U = ( u n k ) ( b s ( F ) , c 0 ) iff Equations (47) and (49) hold and Equation (53) holds with   l = 0 .
(2) U = ( u n k ) ( b s ( F ) , c s 0 ) iff Equations (47)–(50) and Equation (55) hold.
(3) U = ( u n k ) ( b s ( F ) , c ) iff Equations (47)–(50) and Equations (52) and (53) hold.
(4) U = ( u n k ) ( b s ( F ) , c s ) iff Equations (47)–(50) and Equation (56) hold.
(5) U = ( u n k ) ( b s ( F ) , l ) iff Equations (47)–(51) hold.
(6) U = ( u n k ) ( b s ( F ) , b s ) iff Equations (47)–(50) and Equation (54) hold.
(7) U = ( u n k ) ( b s ( F ) , l 1 ) iff Equations (47)–(50) and Equation (58) hold.
(8) U = ( u n k ) ( b s ( F ) , b v ) iff Equations (47)–(50) and Equation (57) hold.
(9) U = ( u n k ) ( b s ( F ) , b v 0 ) iff Equations (57) and (47)–(49) and Equation (51) hold and Equation (53) also holds with   l = 0 .
Corallary 23.
Let us take U = ( u n k ) mentioned in Theorem 19. Then,
(1) U = ( u n k ) ( c s ( F ) , c 0 ) iff Equations (59), (60), Equation (51) hold and Equation (62) also holds with v k = 0 for all k .
(2) U = ( u n k ) ( c s ( F ) , c s 0 ) iff Equations (59), (60), Equation (54) hold and Equation (65) also holds with v k = 0 for all k .
(3) U = ( u n k ) ( c s ( F ) , c ) iff Equations (59), (60), Equation (51) and Equation (62) hold.
(4) U = ( u n k ) ( c s ( F ) , c s ) iff Equations (59), (60), Equation (64) and Equation (65) hold.
(5) U = ( u n k ) ( c s ( F ) , l ) iff Equations (51) and (59)–(61) hold.
(6) U = ( u n k ) ( c s ( F ) , b s ) iff Equations (59), (60), (54) and (63) hold.
(7) U = ( u n k ) ( c s ( F ) , l 1 ) iff Equations (59), (60) and (66) hold.
(8) U = ( u n k ) ( c s ( F ) , b v ) iff Equations (59), (60) and (68) hold.
(9) U = ( u n k ) ( c s ( F ) , b v 0 ) iff Equation (59), (60) and (62) hold and Equation (68) holds with v k = 0 for all k .
Corollary 24.
Let us take U = ( u n k ) mentioned Theorem 19. Then,
(1) U = ( u n k ) ( b s ( F ) , c ^ ) iff Equations (47)–(51), (67) and (69) hold.
(2) U = ( u n k ) ( b s ( F ) , c ^ 0 ) iff Equations (47)–(51) hold and Equations (68) and (69) also hold with v k = 0 in Equation (67) and l k = 0 in Equation (69).
(3) U = ( u n k ) ( c s ( F ) , c ^ ) iff Equations (59), (60), (67) and (51) hold.
(4) U = ( u n k ) ( c s ( F ) , c ^ 0 ) iff Equations (51), (59), (60) hold and Equation (67) also holds with v k = 0 .
(5) U = ( u n k ) ( b s ( F ) , c ^ s ) iff Equations (69), (73), (74) and (47)–(50), hold.
(6) U = ( u n k ) ( c s ( F ) , c ^ s ) iff Equations (73), (74), (59) and (60) hold.
Corallary 25.
Let us take U = ( u n k ) mentioned Theorem 20. Then,
(1) U = ( u n k ) ( l p , b s ( F ) ) iff Equation (40) holds with b n k instead of r n k , where b n k is defined by Equation (31).
(2) U = ( u n k ) ( l , b s ( F ) ) = ( c , b s ( F ) ) = ( c 0 , b s ( F ) ) iff Equation (37) holds with b n k instead of r n k , where b n k is defined by Equation (31).
(3) U = ( u n k ) ( l 1 , b s ( F ) ) iff Equation (41) holds with b n k instead of r n k , where b n k is defined by Equation (31).
(4) U = ( u n k ) ( b v , b s ( F ) ) iff Equation (42) holds with b n k instead of r n k , where b n k is defined by Equation (31).
(5) U = ( u n k ) ( b v 0 , b s ( F ) ) iff Equation (43) holds b n k instead of r n k , where b n k is defined by Equation (31).
(6) U = ( u n k ) ( l , c s ( F ) ) iff Equation (44) holds with b n k instead of r n k , where b n k is defined by Equation (31).
(7) U = ( u n k ) ( c , c s ( F ) ) iff (10), Equations (37) and (47) hold b n k instead of r n k , where b n k is defined by Equation (31).
(8) U = ( u n k ) ( c s 0 , c s ( F ) ) iff Equations (9) and (46) hold with b n k instead of r n k , where b n k is defined by Equation (31).
(9) U = ( u n k ) ( l p , c s ( F ) ) iff Equations (10) and (40) hold with b n k instead of r n k , where b n k is defined by Equation (31).
(10) U = ( u n k ) ( l , c s ( F ) ) iff Equations (10) and (41) hold with b n k instead of r n k , where b n k is defined by Equation (31).
(11) U = ( u n k ) ( b v , c s ( F ) ) iff Equations (10), (41) and (43) hold with b n k instead of r n k , where b n k is defined by Equation (31).
(12) U = ( u n k ) ( b v 0 , c s ( F ) ) iff Equations (10) and (43) hold b n k instead of r n k , where b n k is defined by Equation (31).
(13) U = ( u n k ) ( c ^ , c s ( F ) ) iff Equation (65) and Equations (70)–(72) hold with b n k instead of v n k , where b n k is defined by Equation (31).

3. Discussion

Kızmaz [47] first introduced the difference sequence operator in 1981. Generalized difference sequence spaces were characterized and investigated by Kirişçi and Başar [4] in 2010. Kara [27] first defined the Fibonacci Difference Matrix F, which created the Fibonacci sequence (fn) in 2013. He also introduced the new sequence spaces p(F) and (F); where 1 ≤ p < ∞. The spaces c(F(r,s)) and c0(F(r,s)) were introduced by Candan [28] in 2015. In 2015, the sequence space p(F(r,s)) was introduced and studied by Candan and Kara [19]; where 1 ≤ p ≤ ∞. In addtion, a class of compact operators on p(F) and (F) was characterized by Kara et al. [32], where 1 ≤ p < ∞.
In the present study, we introduced the domain of a triangular infinite matrix on a sequences space. We described spaces cs(F) and bs(F), where F, cs, and bs are the Fibonacci Difference Matrix, convergent and bounded series, respectively. It was demonstrated that bs(F) are the linear spaces, and given that cs(F) is linear space in Theorem 6. without proof and, they have the same norm
x = sup n | j = 0 n ( f k f k + 1 x k f k + 1 f k x k 1 ) |
where xcs(F) or xbs(F). It was found that they are Banach spaces. In addition, inclusions theorems were examined and found. Finally, the γ, β, α -duals of them were calculated. Finally, some matrix transformations as a main result were given.

Author Contributions

This article was written with equal contributions from both authors. The final manuscript was read and approved by both authors.

Funding

This research received no external funding.

Acknowledgments

We thank the responsible editors and reviewers for their contribution for the improvement of this paper. We also thank Mehmet Kanoğlu for his help in reviewing English of the paper.

Conflicts of Interest

The authors declare no conflict of interest.

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Yaşar, F.; Kayaduman, K. On the Domain of the Fibonacci Difference Matrix. Mathematics 2019, 7, 204. https://doi.org/10.3390/math7020204

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Yaşar F, Kayaduman K. On the Domain of the Fibonacci Difference Matrix. Mathematics. 2019; 7(2):204. https://doi.org/10.3390/math7020204

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Yaşar, Fevzi, and Kuddusi Kayaduman. 2019. "On the Domain of the Fibonacci Difference Matrix" Mathematics 7, no. 2: 204. https://doi.org/10.3390/math7020204

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Yaşar, F., & Kayaduman, K. (2019). On the Domain of the Fibonacci Difference Matrix. Mathematics, 7(2), 204. https://doi.org/10.3390/math7020204

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